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Solution of problem 2

Solution 2A. Symmetry described by matrices.

For the problems, see p. .

Answers

(i)

(B,b)(A,a): BA = $\left( \begin{array}{rrr} 1&0&0\\ 0&0&\bar{1}\\ 0&1&0 \end{array} \right)$, Ba = $\left( \begin{array}{r} 1/2\\ 1/2\\ 1/2 \end{array}\right)$,
Ba + b = Ba for b = o.

Therefore, (BA,Ba + b) = (C,c) = $\left( \begin{array}{rrr} 1&0&0\\ 0&0&\bar{1}\\ 0&1&0 \end{array} \right)$, $\left( \begin{array}{r} 1/2\\ 1/2\\ 1/2 \end{array}\right)$.

(ii)

Analogously one calculates

(A,a)(B,b) = (D,d) = $\left( \begin{array}{rrr} 0&0&1\\ 0&1&0\\ \bar{1}&0&0\end{array} \right),\,\left( \begin{array}{r} 1/2\\ 1/2\\ 1/2 \end{array}\right)$.

(iii)

(A,a)$^{\,-1}$ = $\left( \begin{array}{rrr} 0&1&0\\ 1&0&0\\ 0&0&\bar{1} \end{array} \right),\,\... ...n{array}{r}\hspace{-0.3em}-1/2\\ \hspace{-0.3em}-1/2\\ 1/2 \end{array}\right)$.

(iv)

(B,b)$^{\,-1}$ = $\left( \begin{array}{rrr} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array} \right)$, $\left( \begin{array}{r} 0\\ 0\\ 0 \end{array}\right)$.

(v)

(C,c)$^{\,-1}$ = $\left( \begin{array}{rrr} 1&0&0\\ 0&0&1\\ 0&\bar{1}&0 \end{array} \right)$, $\left( \begin{array}{r}\hspace{-0.3em}-1/2\\ \hspace{-0.3em}-1/2 \\ 1/2 \end{array}\right)$.

(vi)

(D,d)$^{\,-1}$ = $\left( \begin{array}{rrr} 0&0&\bar{1}\\ 0&1&0\\ 1&0&0\end{array} \right),\,\l... ...{array}{r} 1/2\\ \hspace{-0.3em}-1/2\\ \hspace{-0.3em}-1/2 \end{array}\right)$.

(vii)

$(\mbox{\textit{\textbf{B,\,b}}})^{\,-1}\,(\mbox{\textit{\textbf{A,\,a}}})^{\,-1... ...x{\textit{\textbf{D,\,d}}})^{\,-1} \ne(\mbox{\textit{\textbf{C,\,c}}})^{\,-1}.$

Note, that (B,b)$^{\,-1}$(A,a)$^{\,-1}$ = [(A,a)(B,b)]$^{\,-1}$ = (D,d)$^{\,-1}$.

Solution 2B. Symmetry described by matrices.

For the problem, see p. .

Answer

(viii)

We follow the procedure described in Section 5.2.

From the matrix parts the `types' of the operations are determined by the determinants and traces:

All the matrices are those of rotations. The directions [uvw] of the rotation axes are determined by applying equation (5.2.3):

It is more or less a matter of taste and experience if one continues with the calculation of the screw part (possibly o) by equation 5.2.5 or if one calculates the (possibly non-existing) fixed points by equation 5.2.7. If the order of the matrix is low then the calculation of the screw part is not so costly as if the order is high. If the screw part turns out to be o or if there are no fixed points then the calculation was not quite in vain because one then knows that the other way will be successful.

`Obviously' the pair (B,b) describes a rotation because the column $\mbox{\textit{\textbf{w}}}=\mbox{\textit{\textbf{o}}}$ indicates the origin $O$ to be a fixed point. Solution: (B,b) describes a 3-fold rotation with rotation axis [111] and the points $x,\,x,\,x$ (including 0,0,0) as fixed points.

We decide to calculate the screw parts in all other cases. Because of the order 2, the calculation for (A,a) is short. The pairs (C,c) and (D,D) can not have fixed points because in both cases a `$+1$' in the main diagonal is combined with a non-zero coefficient $w_i$ in the column. This $w_i$ is a screw coefficient, see the remark on diagonal matrices in Section 5.2. We start with (A,a).

$\frac{1}{2}\hspace{-1mm}\left[\hspace{-0.5mm} \left( \begin{array}{ccc}0&1&0\\ ... ...{-0.5mm}=\hspace{-0.5mm}\left(\begin{array}{c} 1/2\\ 1/2\\ 0\end{array}\right)$

is the screw part of (A,a).

The reduced operation is $(\mbox{\textit{\textbf{A}}},\,\mbox{\textit{\textbf{a}}}_{lp})= \left( \begin{a... ...r{1} \end{array} \right),\left(\begin{array}{c} 0\\ 0\\ 1/2\end{array}\right).$

Equation 5.2.8 yields $y_F=x_F,\ x_F=y_F, -z_F+1/2=z_F$ and the fixed points $x,\,x,\,1/4$, with arbitrary $x$. The fixed points are not really fixed points of the symmetry operation but are the coordinates of the screw-rotation axis $2_1$.

The calculation for (C,c) is a bit more lengthy:

$\frac{\mbox{\textit{\textbf{t}}}}{4}\hspace{-0.5mm}=\hspace{-0.5mm}\frac{1}{4}\... ...ht]\hspace{-0.5mm} \left(\begin{array}{c} 1/2\\ 1/2\\ 1/2 \end{array}\right)=$

= $\frac{1}{4}\left(\begin{array}{ccc}4&0&0\\ 0&0&0\\ 0&0&0 \end{array}\right) ... ...2 \end{array}\right)= \left(\begin{array}{c} 1/2\\ 0\\ 0 \end{array}\right).$

The symmetry operation is a 4-fold screw rotation with $HM$ symbol $4_2$.

The points of the screw axis are determined by equation 5.2.8 again:

$x_F=x_F,\,-z_F+1/2=y_F,\,y_F+1/2=z_F$ result in $x,\,0,\,1/2$ with arbitrary $x$.

Analogously one determines (D,d) to describe a 4-fold screw rotation $4_2$, the screw axis in $1/2,\,y,\,0$, with arbitrary $y$.

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